JEE Mathematics Probability MCQs Set 09

Answer: (b) \( \frac5{54} \)

Explanation:
1. The three numbers must be strictly increasing, so they are three distinct values in increasing order: \( \binom63=20 \) favourable outcomes.
2. Total outcomes \( 6^3=216 \). Probability \( \frac{20}{216}=\frac5{54} \).

Teacher's Note:
1. A strictly increasing sequence is determined by the set of three distinct numbers.
2. Divide by 216.

Answer: (d) none of these

Explanation:
1. Each throw is even with probability 1/2, so the number of even faces is Binomial(2n + 1, 1/2).
2. By the symmetry k ↔ 2n + 1 − k (one is odd, the other even), the probability that the count is odd equals the probability that it is even, which is \( \frac12 \).
3. This is none of (a)-(c).

Teacher's Note:
1. Use the symmetry of the binomial distribution with p = 1/2 and an odd number of trials.
2. P(odd) + P(even) = 1.

Answer: (d) \( ({}^{90}C_3\times{}^{10}C_1)/{}^{100}C_4 \)

Explanation:
1. Choose 3 good bulbs out of the 90 good ones and 1 defective out of the 10 defective ones.
2. Divided by the total ways of choosing 4 of the 100 bulbs: \( \frac{{}^{90}C_3\times{}^{10}C_1}{{}^{100}C_4} \).

Teacher's Note:
1. Favourable selections = good × defective.
2. Use combinations for the total.

Answer: (a) \( \frac15 \)

Explanation:
1. A × B has 5 × 4 = 20 elements.
2. a + b = 9 with a odd and b even: (1, 8), (3, 6), (5, 4), (7, 2): 4 pairs (a = 9 would need b = 0).
3. Probability \( \frac4{20}=\frac15 \).

Teacher's Note:
1. List the favourable pairs explicitly.
2. Check that every b is in B.

Answer: (c) \( \frac{1-P(A\cup B)}{P(\bar B)} \)

Explanation:
1. \( P(\bar A/\bar B)=\frac{P(\bar A\cap\bar B)}{P(\bar B)} \).
2. By De Morgan, \( \bar A\cap\bar B=\overline{A\cup B} \), so the numerator is \( 1-P(A\cup B) \).

Teacher's Note:
1. Use the definition of conditional probability.
2. De Morgan's law converts the intersection of complements.

Answer: (c) \( \frac49 \)

Explanation:
1. \( f'(x)=3x^2+2ax+b\ge0 \) for all x iff \( 4a^2-12b\le0 \), i.e. \( a^2\le3b \).
2. Count (a, b): a = 1: b ≥ 1 (6); a = 2: b ≥ 2 (5); a = 3: b ≥ 3 (4); a = 4: b ≥ 6 (1); a ≥ 5: none. Total 16.
3. With any c (6 ways): \( \frac{16\times6}{216}=\frac49 \).

Teacher's Note:
1. An increasing cubic needs a non-positive discriminant of f′.
2. c is free.

Answer: (b) \( \frac14 \)

Explanation:
1. Let X be the number of tails. By symmetry, \( P(X=k)=P(X=100-k) \).
2. The odd values 1, 3, ..., 49 are mapped to 99, 97, ..., 51, so \( P(X\in\{1,3,\dots,49\})=P(X\in\{51,\dots,99\}\text{ odd}) \).
3. X odd has probability 1/2 (and X = 50 is even), so each half has probability 1/4.

Teacher's Note:
1. Use the symmetry P(X = k) = P(X = 100 − k).
2. Probability of an odd number of tails is 1/2.

Answer: (a) \( \frac5{11} \)

Explanation:
1. The first 1 appears on trial k with probability \( \left(\frac56\right)^{k-1}\frac16 \).
2. Sum over even k = 2, 4, ...: \( \frac16\cdot\frac56\left[1+\frac{25}{36}+\cdots\right]=\frac{5/36}{1-25/36}=\frac5{11} \).

Teacher's Note:
1. Write the probability of the first success at trial k.
2. Sum the geometric series over even k.

Answer: (c) A is true, but R is false.

Explanation:
1. For common difference d (in absolute value), the number of APs inside {1,...,20} is 20 − 3d: d = 1: 17, 2: 14, 3: 11, 4: 8, 5: 5, 6: 2 — a total of 57.
2. Probability \( \frac{57}{\binom{20}4}=\frac{57}{4845}=\frac1{85} \): A is true.
3. R is false because d = 6 also occurs (1, 7, 13, 19).

Teacher's Note:
1. Count APs with each common difference.
2. Check the largest possible difference: 3d ≤ 19 gives d ≤ 6.